# Determinant with Row Multiplied by Constant/Proof 1

## Theorem

Let $\mathbf A = \sqbrk a_n$ be a square matrix of order $n$.

Let $\map \det {\mathbf A}$ be the determinant of $\mathbf A$.

Let $\mathbf B$ be the matrix resulting from one row of $\mathbf A$ having been multiplied by a constant $c$.

Then:

- $\map \det {\mathbf B} = c \map \det {\mathbf A}$

That is, multiplying one row of a square matrix by a constant multiplies its determinant by that constant.

## Proof

Let $\mathbf A = \sqbrk a_n$ be a square matrix of order $n$.

Let $e$ be the elementary row operation that multiplies rows $i$ by the scalar$c$.

Let $\mathbf B = \map e {\mathbf A}$.

Let $\mathbf E$ be the elementary row matrix corresponding to $e$.

From Elementary Row Operations as Matrix Multiplications:

- $\mathbf B = \mathbf E \mathbf A$

From Determinant of Elementary Row Matrix: Exchange Rows:

- $\map \det {\mathbf E} = c$

Then:

\(\ds \map \det {\mathbf B}\) | \(=\) | \(\ds \map \det {\mathbf E \mathbf A}\) | Determinant of Matrix Product | |||||||||||

\(\ds \) | \(=\) | \(\ds c \map \det {\mathbf A}\) | as $\map \det {\mathbf E} = c$ |

Hence the result.

$\blacksquare$

## Sources

- 1998: Richard Kaye and Robert Wilson:
*Linear Algebra*... (previous) ... (next): Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.6$ Determinant and trace: Proposition $1.10$